Curt needs to hit his next shot close to the pin, which is surrounded by water. There is little room for error, and he doesn't know how far his ball is from the pin. Thankfully, Curt has a clever caddy who takes the following actions: walks $50$ meters away from the ball measures the angle between Curt's ball and the pin instructs Curt to measure the angle between him (the caddy) and the pin With this information (shown below), the caddy determines the distance from the ball to the pin. Assuming Curt's clever caddy's calculations are correct, how far is the ball from the pin? Do not round during your calculations. Round your final answer to the nearest meter.
Solution: Converting the problem into geometrical terms Our problem can be modeled by the following triangle $\triangle ABC$, where we want to find $AB=d$. Because the interior angles of a triangle add to $180^\circ$, we know that $\angle A=28^\circ$. $C$ $B$ $69^\circ$ $83^\circ$ $28^\circ$ $A$ $50\text{ m}$ $d$ Since we are given one side length and all angle measures, we can use the law of sines. Using the law of sines $\begin{aligned} \dfrac{\sin(A)}{BC}&=\dfrac{\sin(C)}{AB}\\\\ \dfrac{\sin(28^\circ)}{50} &= \dfrac{\sin(83^\circ)}{d} \gray{\text{Substitute}} \\\\ d \cdot \sin(28^\circ) &= 50 \cdot \sin(83^\circ) \\\\ d &= \dfrac{50 \cdot \sin(83^\circ) }{\sin(28^\circ) } \\\\ d &\approx 106 \,\text{m} \end{aligned}$ The answer The distance from the ball to the pin is $106$ meters.